\subsection{传播子和Feynman公设}
\begin{definition}[][时间演化算符]
    \textbf{time evolution operator}\quad 时间演化算符的物理意义
    是把体系的$t_0$时刻的态矢演化为$t$时刻的态矢.
    $U(t;t_0)$
    \begin{equation*}
        \ket{\psi(t)}=U(t;t_0)\ket{\psi(t_0)}
    \end{equation*}
\end{definition}
如果体系Hamilton量$H$不显含$t$,则从Schrödinger方程可得

\begin{equation*}
    |\psi(t)\rangle=\mathrm{e}^{-\frac{i\left(t-t_0\right) H}{n}}\left|\psi\left(t_0\right)\right\rangle
\end{equation*}
就是说
\begin{equation*}
    U\left(t ; t_0\right)=\mathrm{e}^{-i\left(t-t_0\right) H / \hbar}
\end{equation*}
在坐标表象中,上面态矢的时间演化可具体表述为

\begin{equation*}
    \langle\boldsymbol{r} \mid \psi(t)\rangle=\int\left\langle\boldsymbol{r}\left|U\left(\boldsymbol{t} ; t_0\right)\right| \boldsymbol{r}^{\prime}\right\rangle\left\langle\boldsymbol{r}^{\prime} \mid \psi\left(t_0\right)\right\rangle \mathrm{d} \boldsymbol{r}^{\prime}
\end{equation*}

或为

\begin{equation*}
    \psi(\boldsymbol{r}, t)=\int U\left(\boldsymbol{r}, t ; \boldsymbol{r}^{\prime}, t_0\right) \psi\left(\boldsymbol{r}^{\prime}, t_0\right) \mathrm{d} \boldsymbol{r}^{\prime}
\end{equation*}

这里的积分核

\begin{equation*}
    U\left(\boldsymbol{r}, t ; \boldsymbol{r}^{\prime}, t_0\right)=\left\langle\boldsymbol{r}\left|U\left(t ; t_0\right)\right| \boldsymbol{r}^{\prime}\right\rangle
\end{equation*}

被称为"传播子".它的意思十分清楚地表现在$\psi(\boldsymbol{r}, t)$的积分表示式里,
说明$\psi(\boldsymbol{r}, t)$从何处传播来以及怎样传播来.
若取$\psi\left(\boldsymbol{r}^{\prime}, t_0\right)=\delta\left(\boldsymbol{r}^{\prime}-\boldsymbol{r}_0\right)$,这时

\begin{equation*}
    \psi(\boldsymbol{r}, t)=U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right)
\end{equation*}

这是说,传播子$U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right)$是这样一种概率幅:
粒子在$t_0$时刻位于$r_0$处,演化到$t$时刻位于$r$处的概率幅.
于是,对任意初始的$\psi\left(r_0^{\prime}, t_0\right)$,
在后来的$t$时刻$r$处该粒子的概率幅便可以乘以$\psi\left(r_0^{\prime}, t_0\right)$
并对全部$r_0^{\prime}$积分得到.这里应当强调指出,上面这个传播子$U\left(r, t ; r_0, t_0\right)$
按其物理含义可知,只当$t>t_0$适用.为了这个物理的理由以及数学的理由,
方便的是将其乘以单位阶跃函数$\theta\left(t-t_0\right)$,
就是说,当$t<t_0$时为零.这便是体系的推迟Green函数或称推迟传播子.所以严格说应有

\begin{equation*}
    U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right)=\langle\boldsymbol{r}
    |\mathrm{e}^{-\frac{i}{n}\left(t-t_0\right) \dot{H}}| \boldsymbol{r}_0\rangle
    \theta\left(t-t_0\right)
\end{equation*}

下面在不产生误会的情况下常略写$\theta\left(t-t_0\right)$因子.
在现在路径积分框架里,对有相互作用的一般量子体系,它的传播子$U\left(\boldsymbol{r}, t ; r_0, t_0\right)$
是根据下面Feynman公设给出的:

\begin{definition}[][Feynman公设]
    \textbf{Feynman postulate}\quad
    \begin{enumerate}
        \item 画出连接$\left(r_0, t_0\right)$和$(r, t)$的全部路径$\left(t>t_0\right)$;
        \item 对每条路径$r(t)$,算出作用量$S=\int_4^t L(\boldsymbol{r}(t)) \mathrm{d} \tau$;
        \item $U\left(r, t ; r_0, t_0\right)$即为相因子$\mathrm{e}^{\mathrm{i} S n}$对全部路径求和
              \begin{equation*}
                  U\left(\boldsymbol{r}, t ; r_0, t_0\right)=N \sum_{\text {all of path }} \mathrm{e}^{\mathrm{is} / n}
              \end{equation*}

    \end{enumerate}
\end{definition}
这里$N$是归一化因子.
这里需要对这个公设本身和"全部路径"的含义作一些解释：
设粒子在$t_0$时刻处于$r_0$的概率幅为$\psi\left(r_0, t_0\right)$,经过演化,
到$t$时刻位于$r$处的概率幅为$\psi(r, t)$, Feynman公设主张$\psi(r, t)$按下式决定:

\begin{equation*}
    \psi(\boldsymbol{r}, t)=\int U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right) \psi\left(\boldsymbol{r}_0, t_0\right) \mathrm{d} \boldsymbol{r}_0
\end{equation*}
\begin{figure}[htbp]
    \centering
    \includegraphics[width=0.9\textwidth]{figure/AllPathsOfPathIntegrals20240818220837.jpg}
    \caption{路径积分的全部路径\label{fig:AllPathsOfPathIntegrals20240818220837}}
\end{figure}
"全部路径"的含义可参见\figref{fig:AllPathsOfPathIntegrals20240818220837}.
将$t_0-t$时间分为$n$个等间隔$\varepsilon=t_{i+1}-t_i$,对每一时刻$t_i$ （任意）指定一个位置$r_i$,
这些指定的$\boldsymbol{r}_i(i=1,2, \cdots, n-1)$的集合(加上两个固定的端点
$\left.\boldsymbol{r}_0\right|_{i=0},\left.\boldsymbol{r}\right|_{i=n}$ )便构成一条路径.
重新指定一个、数个或全体$r_i$便又构成了另一条路径.
全部路径的含义是:每一时刻$t_i$所对应的$r_i$均可独立地、任意地改变.
形象地说,粒子在$\left(t_0-t\right)$内将以一切可能画出的路径从$r_0$点到达$\boldsymbol{r}$点.
Feynman公设认为,每条路径对最终概率幅都只贡献一个相因子$\mathrm{e}^{i s / n}$.
严格说来,由于这里的划分是有限区间的划分,仍然不能算是穷尽了全部路径.
只当令$\varepsilon \rightarrow0, n \rightarrow \infty$,取极限,才算是考虑了全部可能的路径.
于是将传播子$U\left(r, t ; r_0, t_0\right)$写出来便成为
\begin{equation*}
    \begin{aligned}
        U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right) & =\lim_{\substack{t \rightarrow0                                                                                                            \\
                n \rightarrow \infty}} \frac{1}{A^3} \int \cdots \int \exp \left\{
        \frac{\mathrm{i} \varepsilon}{\hbar} \sum_{i=0}^{n-1} L\left(
        \frac{\boldsymbol{r}_{i+1}-\boldsymbol{r}_i}{\varepsilon},
        \frac{\boldsymbol{r}_{i+1}+\boldsymbol{r}_i}{2}, \frac{t_{i+1}+t_i}{2}\right)\right\} \frac{\mathrm{d} \boldsymbol{r}_1}{A^3} \cdots \frac{\mathrm{d} \boldsymbol{r}_{n-1}}{A^3}                     \\
                                                                & \equiv \int \exp \left\{\frac{\mathrm{i}}{\hbar} \int_{t_0}^t L[\dot{r}(t), \boldsymbol{r}(t), t] \mathrm{d} t\right\} D \boldsymbol{r}(t)
    \end{aligned}
\end{equation*}


这里$t_n=t, r_n=r$,对取极限的无穷多重数的重积分(泛函积分)采用了特别的标记$\operatorname{Dr}(t)$,
并定义成为的路径积分形式. $A^3$是每重$\mathrm{d} \boldsymbol{r}_i$积分的归一化系数,
它由体系Lagrangian$L$的具体形式决定.当$L=\frac{m}{2} \dot{r}^2-V(r, t)$时, $A$的表达式在下节中给出.
$U$中多出的一个$\frac{1}{A^3}$是考虑到

\begin{equation*}
    \psi(\boldsymbol{r}, t)=\int U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right) \psi\left(r_0, t_0\right) \mathrm{d} r_0
\end{equation*}
表示式里对$\mathrm{d}_0$积分归一化的需要.
\begin{proof}
    在$H$不显含时间的情况下,前面的
    $U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right)=
        \langle\boldsymbol{r}|\mathrm{e}^{-\frac{i}{\hbar}\left(t-t_0\right) \hat{H}}| \boldsymbol{r}_0\rangle$
    表示式和现在用作用量相因子路径积分的表示式是一致的.
    为此考虑$t_i \rightarrow t_{i+1}$这个小区间.注意,在此小区间中路径仍为任意可能的,
    因此即便$t_{i+1}-t_i=\varepsilon$很小, $r_i$也不一定和$\boldsymbol{r}_{i+1}$很接近.这时
    \begin{equation*}
        \begin{aligned}
             & \psi\left(\boldsymbol{r}_{i+1}, t_{i+1}\right)=\int \mathrm{d} \boldsymbol{r}_i\left\langle\boldsymbol{r}_{i+1}\left|\mathrm{e}^{-\frac{i}{n} \varepsilon \dot{H}}\right| \boldsymbol{r}_i\right\rangle \psi\left(\boldsymbol{r}_i, t_i\right)                                                                                                                                                         \\
             & =\iiint \mathrm{d}\left(r_i \boldsymbol{p}_i \boldsymbol{p}_{i+1}\right)\left\langle\boldsymbol{r}_{i+1} \mid \boldsymbol{p}_{i+1}\right\rangle\left\langle\boldsymbol{p}_{i+1}\left|\mathrm{e}^{-\frac{i}{\hbar} \varepsilon \hat{H}}\right| \boldsymbol{p}_i\right\rangle\left\langle\boldsymbol{p}_i \mid \boldsymbol{r}_i\right\rangle \psi\left(\boldsymbol{r}_i, t_i\right)                      \\
             & =\iint \frac{\mathrm{d}\left(\boldsymbol{r}_i \boldsymbol{p}_i \boldsymbol{p}_{i+1}\right)}{(2\pi \hbar)^3} \mathrm{e}^{\frac{\mathrm{i}}{\hbar}\left(p_{i+1} \cdot r_{i+1}-\boldsymbol{p}_i \cdot r_i\right)} \mathrm{e}^{-\frac{i}{\hbar} \varepsilon H\left(p_i \cdot \frac{r_{i+1}+r_i}{2}\right)} \delta\left(\boldsymbol{p}_{i+1}-\boldsymbol{p}_i\right) \psi\left(\boldsymbol{r}_i, t_i\right) \\
             & =\iint \frac{\mathrm{d}\left(\boldsymbol{r}_i \boldsymbol{p}_i\right)}{(2\pi \hbar)^3} \mathrm{e}^{\frac{\mathrm{i}}{\hat{n}} p_i\left(r_{i+1}-r_i\right)-\frac{i}{n} \varepsilon H\left(p_i \cdot \frac{r_{r+1}+r_i}{2}\right)} \psi\left(\boldsymbol{r}_i, t_i\right)                                                                                                                                \\
             & =\int \mathrm{d} \boldsymbol{r}_i \mathrm{e}^{-\frac{i}{n} v v\left(\frac{r_1+r_i}{2}\right)} \psi\left(\boldsymbol{r}_i, t_i\right) \int \frac{\mathrm{d} \boldsymbol{p}_i}{(2\pi \hbar)^3} \mathrm{e}^{\frac{i}{n} p_i\left(r_{r+1}-r_i\right)-\frac{i}{n} \frac{p_i^2}{2m}}
        \end{aligned}
    \end{equation*}
    其中,最后关于$\boldsymbol{p}_i$的积分

    \begin{equation*}
        \begin{aligned}
              & \int \frac{\mathrm{d} \boldsymbol{p}_i}{(2\pi \hbar)^3} \mathrm{e}^{\frac{i}{\hbar} \boldsymbol{p}_i \cdot\left(\boldsymbol{r}_{+1}-\boldsymbol{r}\right)-\frac{i}{\hbar} \varepsilon \frac{p_i^2}{2m}}                                                                                                                                                                                                                             \\
            = & \int \frac{\mathrm{d} \boldsymbol{p}_i}{(2\pi \hbar)^3} \exp \left\{-\frac{\mathrm{i} m}{2\hbar \varepsilon}\left[\frac{\varepsilon^2\boldsymbol{p}_i^2}{m^2}-2\cdot \frac{\varepsilon \boldsymbol{p}_i}{m} \cdot\left(\boldsymbol{r}_{i+1}-\boldsymbol{r}_i\right)+\left(\boldsymbol{r}_{i+1}-\boldsymbol{r}_i\right)^2\right]+\frac{\mathrm{i} m}{2\hbar \varepsilon}\left(\boldsymbol{r}_{i+1}-\boldsymbol{r}_i\right)^2\right\} \\
            = & \frac{1}{(2\pi \hbar)^3} \int \mathrm{d}^3\eta \exp \left\{-\frac{\mathrm{i} m}{2\hbar \varepsilon} \eta^2\right\} \exp \left\{\frac{\mathrm{i} \varepsilon}{\hbar} \frac{m}{2}\left(\frac{\boldsymbol{r}_{i+1}-\boldsymbol{r}_i}{\varepsilon}\right)^2\right\}                                                                                                                                                                     \\
            = & \frac{1}{(2\pi \hbar)^3}\left(\frac{2\pi \hbar \varepsilon}{\mathrm{i} m}\right)^{3/2} \frac{m^3}{\varepsilon^3} \exp \left\{\frac{\mathrm{i} \varepsilon}{\hbar} \frac{m}{2}\left(\frac{\boldsymbol{r}_{i+1}-\boldsymbol{r}_i}{\varepsilon}\right)^2\right\}
        \end{aligned}
    \end{equation*}
    故整个积分
    \begin{equation*}
        \begin{aligned}
            \psi\left(\boldsymbol{r}_{i+1}, t_{i+1}\right) & =\left(\frac{m}{2\pi \hbar \mathrm{i} \varepsilon}\right)^{3/2} \int \mathrm{d} \boldsymbol{r}_i \mathrm{e}^{\frac{i}{\hbar} \varepsilon\left[\frac{m}{2}\left(\frac{r_{i+1}-r_i}{\varepsilon}\right)^2-v\left(\frac{r_{i+1}+r_i}{2}\right)\right]} \psi\left(\boldsymbol{r}_i, t_i\right) \\
                                                           & =\int \frac{\mathrm{d} \boldsymbol{r}_i}{A^3} \mathrm{e}^{\frac{1}{A} \varepsilon L\left(\frac{r_{r+1}-r_i}{\sigma}, \frac{r_{m+1}+r_i}{2}\right)} \psi\left(\boldsymbol{r}_i, t_i\right)
        \end{aligned}
    \end{equation*}

    由这个结果出发，向过去反推至$\left(\boldsymbol{r}_0t_0\right)$ 、
    向将来推至$(\boldsymbol{r} \boldsymbol{t})$ ，并取极限，即得
    \begin{equation*}
        \psi(\boldsymbol{r}, t)=\lim_{\substack{\varepsilon \rightarrow0\\ n \rightarrow \infty}} \int \ldots \int \frac{\mathrm{d} \boldsymbol{r}_{n-1}}{A^3} \ldots \frac{\mathrm{d} \boldsymbol{r}_1}{A^3} \frac{\mathrm{d} \boldsymbol{r}_0}{A^3} \exp \left\{\frac{\mathrm{i} \varepsilon}{\hbar} \sum_{i=0}^{n-1} L\left(\frac{\boldsymbol{r}_{i+1}-\boldsymbol{r}_i}{\varepsilon}, \frac{\boldsymbol{r}_{i+1}+\boldsymbol{r}_i}{2}\right)\right\} \psi\left(\boldsymbol{r}_0, t_0\right)
    \end{equation*}
    与前面
    $\psi(\boldsymbol{r}, t)=\int U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right) \psi\left(\boldsymbol{r}_0, t_0\right) \mathrm{d} \boldsymbol{r}_0$
    相比较可得
    \begin{equation*}
        \begin{aligned}
            U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right) & =\lim_{\substack{t \rightarrow0                                                                                                                                                                                                                                   \\
            n \rightarrow \infty}} \frac{1}{A^3} \int \cdots \int \frac{\mathrm{d} \boldsymbol{r}_{n-1}}{A^3} \cdots \frac{\mathrm{d} \boldsymbol{r}_1}{A^3} \mathrm{e}^{\frac{\mathrm{i} \varepsilon}{\hbar} \sum_{i=0}^{2-1} L}                                                                                                       \\
                                                                    & =\int \exp \left\{\frac{\mathrm{i}}{\hbar} \int_{t_0}^t L(\boldsymbol{r}(\tau)) \mathrm{d} \tau\right\} D \boldsymbol{r}(t)=\left\langle\boldsymbol{r}\left|\mathrm{e}^{-\frac{\mathrm{i}}{\hbar}\left(t-t_0\right) \hat{A}}\right| \boldsymbol{r}_0\right\rangle
        \end{aligned}
    \end{equation*}
    应当指出，传播子的$\mathrm{e}^{i S / \hbar}$路径积分表示式比
    $\mathrm{e}^{-\frac{1}{\hbar}\left(t-t_0\right) \dot{H}}$
    坐标表象矩阵元表示式更为普遍，因为前者对Hamilton量$\hat{H}(t)$含时情况仍适用。
\end{proof}
\subsection{和Schrödinger方程的等价性}

可以证明，上述表示的$\psi(\boldsymbol{r}, t)$满足Schrödinger方程.
为阐述得更为清楚，这里用两种办法来证明.

\begin{proof}
    第一种办法是从路径积分表示式出发,

    \begin{equation*}
        \psi\left(\boldsymbol{r}, t^{\prime}\right)=\int \mathrm{d} \boldsymbol{r}^{\prime} U\left(\boldsymbol{r}, t^{\prime} ; \boldsymbol{r}^{\prime}, t\right) \psi\left(\boldsymbol{r}^{\prime}, t\right)
    \end{equation*}


    这里$U$用路径积分表示式.假设
    $t^{\prime}=t+\varepsilon, \varepsilon \rightarrow0, \boldsymbol{r}^{\prime}=\boldsymbol{r}+\boldsymbol{\eta}$
    （注意$\boldsymbol{\eta}$不一定很小，如前所说),这时可取一重积分,即有

    \begin{equation*}
        \begin{aligned}
            \psi(\boldsymbol{r}, t+\varepsilon) & =\int \frac{\mathrm{d} \boldsymbol{r}^{\prime}}{A^3} \exp \left\{\frac{\mathrm{i} \varepsilon}{\hbar}\left[\frac{m}{2}\left(\frac{\boldsymbol{r}-\boldsymbol{r}^{\prime}}{\varepsilon}\right)^2-V\left(\frac{\boldsymbol{r}+\boldsymbol{r}^{\prime}}{2}, \frac{t+t^{\prime}}{2}\right)\right]\right\} \psi(\boldsymbol{r}, t) \\
                                                & =\int \frac{\mathrm{d} \boldsymbol{\eta}}{A^3} \exp \left\{\frac{\mathrm{i} \varepsilon}{\hbar}\left[\frac{m}{2} \frac{\eta^2}{\varepsilon^2}-V\left(\boldsymbol{r}+\frac{\eta}{2}, t+\frac{\varepsilon}{2}\right)\right]\right\} \psi(\boldsymbol{r}+\boldsymbol{\eta}, t)
        \end{aligned}
    \end{equation*}
    由于第一个指数中包含$\frac{\eta^2}{\varepsilon}$,当$\eta^2$不是很小时,这项指数将快速振荡,
    从而对$\boldsymbol{\eta}$积分贡献非常小；所以仅当$\eta^2$是$\frac{\varepsilon \hbar}{m}$量级的小量时，
    第一个指数相位改变不大于1rad量级，积分的主要贡献正是来源于这个量级的$\eta^2$值。
    于是，如果整个等式展开保留到$\varepsilon$量级的项， $\psi$展开只需保留到$\varepsilon$量级的项，
    而对$\boldsymbol{\eta}$展开则要保留到$\eta^2$量级的项

    \begin{equation*}
        \begin{aligned}
            \psi(\boldsymbol{r}, t)+\varepsilon \frac{\partial \psi(\boldsymbol{r}, t)}{\partial t}= & \int \frac{\mathrm{d} \eta}{A^3} \mathrm{e}^{\frac{\mathrm{i} m \eta^2}{2\hbar r}}\left(1-\frac{\mathrm{i} \varepsilon}{\hbar} V(\boldsymbol{r}, t)\right)\left[\psi(\boldsymbol{r}, t)+\boldsymbol{\eta} \cdot \frac{\partial \psi(\boldsymbol{r}, t)}{\partial \eta}\right. \\
                                                                                                     & \left.+\frac{1}{2}\left(\eta_x^2\frac{\partial^2\psi}{\partial x^2}+\eta_y^2\frac{\partial^2\psi}{\partial y^2}+\eta_x^2\frac{\partial^2\psi}{\partial z^2}\right)\right]
        \end{aligned}
    \end{equation*}
    这里，鉴于$V$前面已乘有$\varepsilon ， V$中参量可取零阶近似.利用数学公式（以$\eta_x$情况为例）
    \begin{equation*}
        \int_{-\infty}^{+\infty} \mathrm{d} \eta_x \mathrm{e}^{\frac{\mathrm{i} m \eta_x^2}{2\lambda \varepsilon}}=A=\left(\frac{\mathrm{i}2\pi \hbar \varepsilon}{m}\right)^{1/2}, \quad \int_{-\infty}^{+\infty} \frac{\mathrm{d} \eta_x}{A} \eta_x \mathrm{e}^{\frac{\mathrm{i} m \eta_x^2}{2h \varepsilon}}=0
    \end{equation*}
    上式变成
    \begin{equation*}
        \int_{-\infty}^{+\infty} \frac{\mathrm{d} \eta_x}{A} \eta_x^2\mathrm{e}^{\frac{\mathrm{im} \eta_x^2}{2\lambda_x}}=\frac{\mathrm{i} \hbar \varepsilon}{m}
    \end{equation*}
    (对第一个积分等式两边$\varepsilon$作微分,即得此等式)即
    \begin{equation*}
        \begin{gathered}
            \psi+\varepsilon \frac{\partial \psi}{\partial t}=\psi-\frac{i \varepsilon}{\hbar} V \psi+\frac{\mathrm{i} \hbar \varepsilon}{2m} \Delta \psi \\
            \mathrm{i} \hbar \frac{\partial \psi}{\partial t}=\left(-\frac{\hbar^2}{2m} \Delta+V\right) \psi
        \end{gathered}
    \end{equation*}
\end{proof}
\begin{proof}
    第二种办法是从
    \begin{equation*}
        \psi\left(\boldsymbol{r}, t^{\prime}\right)=\int\left\langle\boldsymbol{r}\left|\mathrm{e}^{-\mathrm{i}\left(t^{\prime}-t\right) \hat{H} / \hbar}\right| \boldsymbol{r}_0\right\rangle \psi\left(\boldsymbol{r}_0, t\right) \mathrm{d} \boldsymbol{r}_0
    \end{equation*}
    出发.现假定$t^{\prime}=t+\varepsilon$,于是上式成为
    \begin{equation*}
        \begin{aligned}
            \psi(\boldsymbol{r}, t+\varepsilon) & =\int\left\langle\boldsymbol{r}\left|\mathrm{e}^{-\mathrm{i} \frac{\varepsilon}{\hbar} \hat{H}}\right| \boldsymbol{r}_0\right\rangle \psi\left(\boldsymbol{r}_0, t\right) \mathrm{d} \boldsymbol{r}_0  \\
                                                & =\int\left\langle\boldsymbol{r}\left|\left(1-\mathrm{i} \frac{\varepsilon}{\hbar} \hat{H}\right)\right| \boldsymbol{r}_0\right\rangle \psi\left(\boldsymbol{r}_0, t\right) \mathrm{d} \boldsymbol{r}_0
        \end{aligned}
    \end{equation*}
    注意到
    \begin{equation*}
        \begin{aligned}
            \int \mathrm{d} r_0\left\langle\boldsymbol{r}|\hat{\boldsymbol{p}}| \boldsymbol{r}_0\right\rangle \psi\left(\boldsymbol{r}_0, t\right) & =\int \mathrm{d} \boldsymbol{r}_0\left(-\mathrm{i} \hbar \frac{\partial}{\partial \boldsymbol{r}}\left\langle\boldsymbol{r} \mid \boldsymbol{r}_0\right\rangle\right) \psi\left(\boldsymbol{r}_0, t\right)                                                      \\
                                                                                                                                                   & =-\mathrm{i} \hbar \frac{\partial}{\partial \vec{r}} \int \mathrm{d} \boldsymbol{r}_0\delta\left(\boldsymbol{r}-\boldsymbol{r}_0\right) \psi\left(\boldsymbol{r}_0, t\right)=-\mathrm{i} \hbar \frac{\partial}{\partial \boldsymbol{r}} \psi(\boldsymbol{r}, t)
        \end{aligned}
    \end{equation*}
    代入上式即得
    \begin{equation*}
        \psi(\boldsymbol{r}, t+\varepsilon)=\psi(\boldsymbol{r}, t)-\frac{\mathrm{i} \varepsilon}{\hbar} \hat{H}\left(-\mathrm{i} \hbar \frac{\partial}{\partial \vec{r}}, \boldsymbol{r}\right) \psi(\boldsymbol{r}, t)
    \end{equation*}
    也即
    \begin{equation*}
        \mathrm{i} \hbar \frac{\partial \psi}{\partial t}=\hat{H} \psi
    \end{equation*}
\end{proof}

注意这时也给出了动量算符在坐标表象中的表达式,于是也就同时给出了基本对易规则.路径积分方法的这种能力将来十分有用.
\subsection{传播子再研究}
其实,上述传播子即为Schrödinger方程的Green函数.下面导出它所满足的微分方程给以证实.
\begin{equation*}
    \begin{aligned}
          & \frac{\partial}{\partial t_2}\left\{\left\langle r_2\left|\mathrm{e}^{-i\left(t_2-r_1\right) \hat{A} / n}\right| \boldsymbol{r}_1\right\rangle \theta\left(t_2-t_1\right)\right\}                                                                                                                                                                          \\
        = & -\frac{1}{\hbar}\left\langle\boldsymbol{r}_2\left|\hat{H} \mathrm{e}^{-\frac{i}{\hbar}\left(t_2-t_1\right) \hat{A} / \hbar}\right| r_1\right\rangle \cdot \theta\left(t_2-t_1\right)+\left\langle\boldsymbol{r}_2\left|\mathrm{e}^{-i\left(t_2-t_4\right) \dot{H} / n}\right| r_1\right\rangle \delta\left(t_2-t_1\right)                                  \\
        = & -\frac{\mathrm{i}}{\hbar} \hat{H}_2\left\langle\boldsymbol{r}_2\left|\mathrm{e}^{-\frac{i}{\hbar}\left(t_2-t_1\right) \hat{H}}\right| \boldsymbol{r}_1\right\rangle \theta\left(t_2-t_1\right)+\left\langle\boldsymbol{r}_2\left|\mathrm{e}^{-\frac{1}{\hbar}\left(t_2-t_2\right) \hat{H}}\right| \boldsymbol{r}_1\right\rangle \delta\left(t_2-t_1\right)
    \end{aligned}
\end{equation*}
这里用了$\left\langle\boldsymbol{r}_2\right| \hat{\boldsymbol{p}}=-\mathrm{i} \hbar \nabla_2\left\langle\boldsymbol{r}_2\right| 、 \hat{H}_2(\hat{\boldsymbol{p}}, \hat{\boldsymbol{r}})=\hat{H}_2\left(-\mathrm{i} \hbar \nabla_2, \boldsymbol{r}_2\right)$,
将算符$\hat{\boldsymbol{p}}$转化为对$\boldsymbol{r}_2$的作用.
同时,由于第二项含有$\delta\left(t_2-t_1\right)$因子,可令该项矩阵元内指数上$t_2=t_1$,于是可得

\begin{equation*}
    \mathrm{i} \hbar \frac{\partial U(2;1)}{\partial t_2}-\hat{H}_2U(2;1)=\mathrm{i} \hbar \delta\left(\boldsymbol{r}_2-\boldsymbol{r}_1\right) \delta\left(t_2-t_1\right)
\end{equation*}

这里已简记$U\left(\boldsymbol{r}_2, t_2; \boldsymbol{r}_1, t_1\right)=U(2;1)$.
这就是$U(2;1)$作为量子体系推迟Green函数所满足的微分方程.严格地说,此方程再加辅助条件

\begin{equation*}
    U\left(\boldsymbol{r}_2, t_2; \boldsymbol{r}_1, t_1\right)=0\quad\left(t_2<t_1\right)
\end{equation*}

才可以完全决定所要的推迟Green函数.
同样,若$H$不显含时间,将有定态解完备集合

\begin{equation*}
    \left\{\begin{array}{l}
        H\left|\varphi_n\right\rangle=E_n\left|\varphi_n\right\rangle \\
        \sum_n\left|\varphi_n\right\rangle\left\langle\varphi_n\right|=1
    \end{array}\right.
\end{equation*}
于是这时传播子可明显地写为
\begin{equation*}
    \begin{aligned}
        U(2;1) & =\left\langle r_2\left|\mathrm{e}^{-\frac{i}{\hbar}\left(t_2-t_1\right) \hat{H}}
        \right| r_1\right\rangle \theta\left(t_2-t_1\right)                                                                                                             \\
               & =\sum_{m n}\left\langle r_2\mid \varphi_m\right\rangle\left\langle\left.\varphi_m
        \mathrm{e}^{-\frac{i}{\hbar}\left(t_3-t_1\right) \hat{H}} \right|, \varphi_n\right\rangle\left\langle\varphi_n \mid r_1\right\rangle \theta\left(t_2-t_1\right) \\
               & =\sum_n \varphi_n^*\left(r_1\right) \varphi_n\left(r_2\right)
        \mathrm{e}^{-\frac{i}{\hbar}\left(t_2-t_1\right) \hat{H}} \theta\left(t_2-t_1\right)
    \end{aligned}
\end{equation*}
另外,由于时间上相继发生的事件,其概率幅将相乘,于是将所有的路径进行两次或多次分割是完全可能的.比如,这时将有

\begin{equation*}
    U(3;1)=\int U(3;2) U(2;1) \mathrm{d} r_2
\end{equation*}

这从路径积分思想或直接在表达式中插入坐标表象基矢完备性条件均可看出.
最后,应当指出,还有另一种很常用的求积路径积分的办法,这就是将端点在$\left(x_0, t_0\right)$和$(x, t)$
的任意路径分解为经典路径及量子涨落两部分
\begin{equation*}
    x(t)=x_{\mathrm{C}}(t)+y(t)
\end{equation*}
相应地,作用量也被区分开来.设$S_{\mathrm{c}}$是Lagrangian $L$沿着经典允许轨道从
$\left(x_0, t_0\right)$到$(x, t)$的时间积分.可以证明有
\begin{equation*}
    U\left(x, t ; x_0, t_0\right)=\exp \left\{\frac{\mathrm{i}}{\hbar} S_{\mathrm{C}}\left(x, t ; x_0, t_0\right)\right\} F\left(t ; t_0\right)
\end{equation*}
涨落因子$F$由对$y(t)$的路径积分而得,只依赖于两端点时间,与空间变数无关。
\subsection{路径积分计算——自由粒子}

下面用三种方法计算自由粒子的传播子.
第一,直接积分计算.这时$H=\frac{p^2}{2m}$,演化算符为

\begin{equation*}
    U\left(t, t_0\right)=\int \mathrm{e}^{-i\left(t-t_0\right) \frac{p^2}{2m \hbar}}|p\rangle\langle p| \mathrm{d} p
\end{equation*}
传播子为
\begin{equation*}
    \begin{aligned}
        U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right)
         & =\left\langle\boldsymbol{r}\left|U\left(t, t_0\right)\right| \boldsymbol{r}_0\right\rangle=\int \mathrm{e}^{-i \frac{\left(t-t_0\right) p^2}{2m n}}\langle\boldsymbol{r} \mid \boldsymbol{p}\rangle\left\langle\boldsymbol{p} \mid \boldsymbol{r}_0\right\rangle \mathrm{d} \boldsymbol{p} \\
         & =\frac{1}{(2\pi \hbar)^3} \mathrm{e}^{-\mathrm{i} \frac{\left(t-t_0\right) p^2}{2m \hbar}+\frac{i\left(-\left(r-r_0\right)\right.}{\hbar}} \mathrm{d} \boldsymbol{p}                                                                                                                       \\
         & =\left(\frac{m}{i2\pi \hbar\left(t-t_0\right)}\right)^{3/2} \exp \left(\frac{\mathrm{i} m}{2\hbar\left(t-t_0\right)}\left(\boldsymbol{r}-\boldsymbol{r}_0\right)^2\right) \theta\left(t-t_0\right)
    \end{aligned}
\end{equation*}
可以直接检验,当$t>t_0$时,这个传播子满足Schrödinger方程.

第二,用标准的路径积分方法,即逐重积分进行计算.这时$L=\frac{m}{2} \dot{\boldsymbol{r}}^2$,

\begin{equation*}
    U\left(\boldsymbol{r}, t ; \boldsymbol{r}_0, t_0\right)=\lim_{\substack{n \rightarrow \infty \\ \varepsilon \rightarrow0}}\left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{-\frac{3}{2}} \int \cdots \int \frac{\mathrm{d} \boldsymbol{r}_{n-1}}{\left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{3/2}} \cdots \frac{\mathrm{d} \boldsymbol{r}_1}{\left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{3/2}} \exp \left\{\frac{\mathrm{i} m}{2\hbar \varepsilon} \sum_{i=0}^{n-1}\left(\boldsymbol{r}_{i+1}-\boldsymbol{r}_i\right)^2\right\}
\end{equation*}
这里$\boldsymbol{r}_n=\boldsymbol{r}, n \varepsilon=t-t_0$.由于这时
$x , y , z$三个自由度彼此独立,计算完全相同,为书写简单只对$x$自由度进行这个多重积分计算.
先从对$x_1$的积分开始,将这个积分乘以
$\left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{-\frac{1}{2}}$
(此因子原在重积分号之外),即得

\begin{equation*}
    \begin{aligned}
          & \left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{-1} \int_{-\infty}^{+\infty} \mathrm{d} x_1\exp \left\{\frac{\mathrm{i} m}{2\hbar \varepsilon}\left[\left(x_2-x_1\right)^2+\left(x_1-x_0\right)^2\right]\right\}                        \\
        = & \left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{-1} \int_{-\infty}^{+\infty} \mathrm{d} x_1\exp \left\{\frac{\mathrm{i} m}{\hbar \varepsilon}\left[\left(x_1-\frac{x_2+x_0}{2}\right)^2+\frac{\left(x_2-x_0\right)^2}{4}\right]\right\}
    \end{aligned}
\end{equation*}

利用广义Gauss积分公式，上式即为

\begin{equation*}
    \left(\frac{\pi \mathrm{i} \cdot2\hbar \cdot2\varepsilon}{m}\right)^{-\frac{1}{2}} \exp \left\{\frac{\mathrm{i} m}{2\hbar \cdot2\varepsilon}\left(x_2-x_0\right)^2\right\}
\end{equation*}

接着将此结果乘以

\begin{equation*}
    \left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{-\frac{1}{2}} \exp \left\{\frac{\mathrm{i} m}{2\hbar \varepsilon}\left(x_3-x_2\right)^2\right\}
\end{equation*}
再进行$x_2$积分

\begin{equation*}
    \begin{aligned}
          & \frac{m}{\pi \mathrm{i} \sqrt{2} \hbar \cdot2\varepsilon} \int_{-\infty}^{+\infty} \mathrm{d} x_2\exp \left\{\frac{\mathrm{i} m}{4\hbar \varepsilon}\left(x_2-x_0\right)^2+\frac{\mathrm{i} m}{2\hbar \varepsilon}\left(x_3-x_2\right)^2\right\}                               \\
        = & \frac{m}{\pi \mathrm{i} \sqrt{2} \hbar \cdot2\varepsilon} \int_{-\infty}^{+\infty} \mathrm{d} x_2\exp \left\{\frac{\mathrm{i} m}{4\hbar \varepsilon}\left[\left(\sqrt{3} x_2-\frac{1}{\sqrt{3}}\left(2x_3+x_0\right)\right)^2+\frac{2}{3}\left(x_3-x_0\right)^2\right]\right\} \\
        = & \left(\frac{\pi \mathrm{i} \cdot2\hbar \cdot3\varepsilon}{m}\right)^{-\frac{1}{2}} \exp \left\{\frac{\mathrm{i} m}{2\hbar \cdot3\varepsilon}\left(x_3-x_0\right)^2\right\}
    \end{aligned}
\end{equation*}

由此重复并递推下去,直至完成对$x_{n-1}$的积分,将得到
\begin{equation*}
    \left(\frac{\pi \mathrm{i} \cdot2\hbar \cdot n \varepsilon}{m}\right)^{-\frac{1}{2}} \exp \left\{\frac{\mathrm{i} m}{2\hbar \cdot n \varepsilon}\left(x-x_0\right)^2\right\}
\end{equation*}

注意$n \varepsilon=t-t_0$,于是得到$U\left(r, t ; r_0, t_0\right)$的$x$方向部分.
$y , z$计算及结果类似.证毕.

第三,用$n$维重积分的下述公式作一次性积分,给出结果.已知公式

\begin{equation*}
    \int_{-\infty}^{+\infty} \exp \left\{-\sum_{i, j=1}^n M_{i j} x_i x_j+\sum_{i=1}^n \alpha_i x_i\right\} \mathrm{d} x_1\cdots \mathrm{d} x_n=\sqrt{\frac{\pi^n}{\operatorname{det} M}} \mathrm{e}^{\frac{1}{4} \tilde{4} M^{-1} \alpha}
\end{equation*}
这里积分收敛的充分条件要求复对称矩阵$M$满秩并且实部至少为半正定.对自由粒子情况,
$x=\left(x_1, \cdots, x_{n-1}\right), M=-\frac{\mathrm{i} m}{2\hbar \varepsilon} M^{\prime}$
为$(n-1)$维对称矩阵,而

\begin{equation*}
    M^{\prime}=\left(\begin{array}{ccccc}
            2  & -1 &        &    & 0  \\
            -1 & 2  &        &    &    \\
               &    & \ddots &    &    \\
               &    &        & 2  & -1 \\
            0  &    &        & -1 & 2
        \end{array}\right), \quad \tilde{\alpha}=-\frac{\mathrm{i} m}{\hbar \varepsilon} \underbrace{\left(x_0,0, \cdots,0, x\right)}_{(n-1)}
\end{equation*}
可用归纳法验算,
$\operatorname{det} M=\left(-\frac{\mathrm{i} m}{2\hbar \varepsilon}\right)^{n-1} n$;
其逆矩阵为

\begin{equation*}
    M^{-1}=\frac{1}{\operatorname{det} M}\left((-1)^{i+j} \operatorname{det} M^{(i, j)}\right)^{\mathrm{T}}
\end{equation*}

这里$M^{(i, j)}$是从$(n-1)$阶$M$矩阵中删除第$i$行、第$j$列之后的$(n-2)$阶矩阵;
$T$是对括号中矩阵元为$(-1)^{i+j} \operatorname{det} M^{(i, j)}$的$(n-1)$阶矩阵的转置.
于是可得$M^{-1}$的4个有关的矩阵元为

\begin{equation*}
    \begin{gathered}
        \left(M^{-1}\right)_{1,1}=\left(M^{-1}\right)_{n-1, n-1}=\left(-\frac{\mathrm{i} m}{2\hbar \varepsilon}\right)^{-1} \frac{n-1}{n} \\
        \left(M^{-1}\right)_{1, n-1}=\left(M^{-1}\right)_{n-1,1}=\left(-\frac{\mathrm{i} m}{2\hbar \varepsilon}\right)^{-1} \frac{1}{n}
    \end{gathered}
\end{equation*}

后面两个矩阵元相等是由于对称矩阵的逆仍是对称矩阵.
注意,除去第1行第$(n-1)$列后的$(n-2)$阶矩阵$M^{((1, n-1)}$的行列式为
$(-1)^{n-2}$.将以上结果代入重积分公式,得
\begin{equation*}
    \begin{aligned}
        U\left(x, t ; x_0, t_0\right)= & \lim_{\substack{n \rightarrow \infty                                                                                                                                                                                                                \\
        \varepsilon \rightarrow0}} \frac{1}{\left(\frac{2\pi \mathrm{i} \hbar \varepsilon}{m}\right)^{n /2}} \sqrt{\frac{\pi^{n-1}}{\left(-\mathrm{i} \frac{m}{2\hbar \varepsilon}\right)^{n-1} n}}                                                                                          \\
                                       & \cdot \exp \left\{\frac{\mathrm{i} m}{2\hbar \varepsilon}\left(x^2+x_0^2\right)+\frac{1}{4}\left(-\frac{\mathrm{i} m}{\hbar \varepsilon}\right)^2\cdot\left(x_0,0, \cdots,0, x\right) M^{-1}\left(\begin{array}{c}
                                                                                                                                                                                                                                                   x_0    \\
                                                                                                                                                                                                                                                   0      \\
                                                                                                                                                                                                                                                   \vdots \\
                                                                                                                                                                                                                                                   0      \\
                                                                                                                                                                                                                                                   x
                                                                                                                                                                                                                                               \end{array}\right)\right\}                        \\
        =                              & \lim_{\substack{n \rightarrow \infty                                                                                                                                                                                                                \\
        \varepsilon \rightarrow0}}\left(\frac{m}{\mathrm{i}2\pi \hbar n \varepsilon}\right)^{1/2} \exp \left\{\frac{\mathrm{i} m}{2\hbar \varepsilon}\left(x^2+x_0^2\right)-\frac{\mathrm{i} m}{2\hbar \varepsilon}\left[\left(x_0^2+x^2\right) \frac{n-1}{n}+\frac{2x_0x}{n}\right]\right\} \\
        =                              & \left(\frac{m}{\mathrm{i}2\pi \hbar\left(t-t_0\right)}\right)^{1/2} \exp \left\{\frac{\mathrm{i} m}{2\hbar\left(t-t_0\right)}\left(x-x_0\right)^2\right\}
    \end{aligned}
\end{equation*}

\subsection{路径积分计算一一谐振子}

由于已经知道谐振子的本征值和本征函数,采用前面(5.56)式可以更简捷地计算出谐振子的传播子.由前面叙述可知

\begin{equation*}
    U\left(x t ; x_0t_0\right)=\sum_n \varphi_n(x) \varphi_n^*\left(x_0\right) \mathrm{e}^{-\frac{i}{\hbar}\left(t-t_0\right) E_n} \quad\left(t>t_0\right)
\end{equation*}

这时$E_n=\left(n+\frac{1}{2}\right) \hbar \omega$,并且

\begin{equation*}
    \varphi_n(x)=\left(2^n n!\right)^{-1/2}\left(\frac{m \omega}{\pi \hbar}\right)^{1/4} \mathrm{H}_n\left(x \sqrt{\frac{m \omega}{\hbar}}\right) \mathrm{e}^{-\frac{m \omega}{2\hbar} x^2}
\end{equation*}

于是

\begin{equation*}
    U\left(x t ; x_0t_0\right)=\sqrt{\frac{m \omega}{\pi \hbar}} \mathrm{e}^{-\frac{m \omega}{2n}\left(x^2+x_0^2\right)-\frac{i}{2} \omega\left(t-t_0\right)} \sum_{n=0}^{\infty}\left(2^n n!\right)^{-1} \mathrm{H}_n\left(x \sqrt{\frac{m \omega}{\hbar}}\right) \mathrm{H}_n\left(x_0\sqrt{\frac{m \omega}{\hbar}}\right) \mathrm{e}^{-i n \omega\left(t-t_0\right)}
\end{equation*}

利用厄米多项式的积分表示

\begin{equation*}
    \mathrm{H}_n(\xi)=\frac{1}{\sqrt{\pi}} \mathrm{e}^{\xi^2} \int_{-\infty}^{+\infty}\left(\frac{2}{\mathrm{i}}\right)^n \mathrm{e}^{-\tau^2+2\xi \mathrm{ir}} \tau^n \mathrm{~d} \tau
\end{equation*}

得
\begin{equation*}
    \begin{aligned}
        U\left(x t ; x_0t_0\right) & =\sqrt{\frac{m \omega}{\pi^3\hbar}} \mathrm{e}^{\frac{m \omega}{2h}\left(x^2+x_0^2\right)-\frac{1}{2} \omega\left(t-t_0\right)} \iint_{-\infty}^{+\infty} \mathrm{d} \tau \mathrm{d} \tau^{\prime}                                                                                         \\
                                   & \cdot \exp \left\{-\tau^2-\tau^{\prime2}+2x \sqrt{\frac{m \omega}{\hbar}} \mathrm{i} \tau+2x_0\sqrt{\frac{m \omega}{\hbar}} \mathrm{i} \tau^{\prime}\right\} \sum_{n=0}^{\infty} \frac{(-1)^n}{n!}\left(2\tau \tau^{\prime}\right)^n \mathrm{e}^{- \text {ino }\left(t-t_0\right)}         \\
                                   & =\sqrt{\frac{m \omega}{\pi^3\hbar}} \mathrm{e}^{\frac{m \omega}{2\pi}\left(x^2+x_0^2\right)-\frac{1}{2} \omega\left(t-t_0\right)} \int_{-\infty}^{+\infty} \mathrm{d} \tau \mathrm{e}^{-\tau^2+2x \sqrt{\frac{m \omega}{n}}} \mathrm{ir} \int_{-\infty}^{+\infty} \mathrm{d} \tau^{\prime} \\
                                   & \cdot \exp \left\{-\tau^{\prime2}+2x_0\sqrt{\frac{m \omega}{\hbar}} \mathrm{i} \tau^{\prime}-2\tau \tau^{\prime} \mathrm{e}^{-\mathrm{i} \omega\left(t-\epsilon_0\right)}\right\}
    \end{aligned}
\end{equation*}
对$\tau^{\prime}$和$\tau$的积分均可归结为如下积分:

\begin{equation*}
    \int_{-\infty}^{+\infty} \mathrm{e}^{-y^2} \mathrm{~d} y=\sqrt{\pi}
\end{equation*}

完成积分之后得

\begin{equation*}
    \begin{aligned}
        U\left(x t ; x_0t_0\right)= & \sqrt{\frac{m \omega}{\pi \hbar}} \mathrm{e}^{\frac{m \omega}{2}\left(x^2-x_0^2\right)-\frac{i}{2} \omega\left(t-t_0\right)}\left(1-\mathrm{e}^{-2i \omega\left(t-t_0\right)}\right)^{-\frac{1}{2}}                                                 \\
                                    & \cdot \exp \left\{-\frac{m \omega}{\hbar}\left[\frac{x^2-2x x_0\mathrm{e}^{-i \omega\left(t-t_0\right)}+x_0^2\mathrm{e}^{-2i \omega\left(t-t_0\right)}}{1-\mathrm{e}^{-2i \omega \omega\left(t-t_0\right)}}\right]\right\}                          \\
        =                           & \left(\frac{m \omega}{2\pi \mathrm{i} \hbar \sin \omega\left(t-t_0\right)}\right)^{1/2} \exp \left\{\frac{\mathrm{i} m \omega}{2\hbar \sin \omega\left(t-t_0\right)}\left[\left(x^2+x_0^2\right) \cos \omega\left(t-t_0\right)-2x_0x\right]\right\}
    \end{aligned}
\end{equation*}

如果用路径积分方法直接得到此结果,接着再继以此处计算的逆过程,
便成为在路径积分框架下求解简谐振子本征值和本征函数过程,
正如Feynman所做的.
